Question

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Let P be a polynomial function, and P(x) = x^4 - dx^3 + 8x^2 - 14x + 16. If (x-2) is a factor of the polynomial, what is the value of d?​
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Answer 1

Answer:

d = 4.5

Step-by-step explanation:

The given polynomial is P(x) = x⁴ - d·x³ + 8·x² - 14·x + 16

The factor of the polynomial is (x - 2)

By the factor theorem, if (x - 2) is a factor, then P(x) = 0 at x = 2

Therefore, we have;

P(2) = 2⁴ - d·2³ + 8·2² - 14·2 + 16 = 0

2⁴ - d·2³ + 8·2² - 14·2 + 16 = 0

16 - 8·d + 8 × 4 - 28 + 16 = 0

8·d = 16 + 8 × 4 - 28 + 16 = 36

8·d = 36

d = 36/8 = 4.5

d = 4.5

Answer 2

The value of d which makes (x-2) a factor of the polynomial, P(x) is; d = -4.5

If (x-2) is a factor of the polynomial;

As such, x = 2 is a zero of the polynomial function;

In essence, at x = 2, P(x) = 0.

• P(x) = 2⁴ + d(2)³ + 8(2)² - 14(2) + 16 = 0

• 16 + 8d + 32 - 28 + 16 = 0

• 8d +36 = 0.

• 8d = -36

• d = -36/8

d = -4.5

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